## More Lie algebra constructions July 28, 2009

Posted by Akhil Mathew in algebra, representation theory.
Tags: , , ,

The ultimate aim in the series on Lie algebras I am posting here is to cover the representation theory of semisimple Lie algebras. To get there, we first need to discuss some technical tools—for instance, invariant bilinear forms.

Generalities on representations

Fix a Lie algebra ${L}$. Given representations ${V_1, V_2}$, we clearly have a representation ${V_1 \oplus V_2}$; given a morphism of representations ${V_1 \rightarrow V_2}$, i.e. one which respects the action of ${L}$, the kernel and image are themselves representations.

Proposition 1 The category ${Rep(L)}$ of finite-dimensional representations of ${L}$ is an abelian category.

There are a couple of easy technical facts to check (e.g. a monomorphism is the kernel of its cokernel), but we have essentially proved this by the above discussion.

We know, by the embedding theorem, that a small abelian category is a full subcategory of the category ${Mod(R)}$ of left ${R}$-modules for some non-unique ring ${R}$. In our case, we could take ${R = \mathcal{U}(L)}$.

There are a few other ways we can build new representations from old ones:

Definition 2 If ${V, W}$ are representations of ${L}$, then ${V \otimes W}$ is a representation under the action $\displaystyle X(v \otimes w) := (Xv) \otimes w + v \otimes (Xw).\ \ \ \ \ (1)$

The fact that this is a representation can be directly checked. Moreover, the reason for this somewhat odd choice comes from the theory of Lie groups: If the Lie group ${G}$ acts on ${V, W}$, then it acts on ${V \otimes W}$ by $\displaystyle g(v \otimes w) := gv \otimes gw,$

and differentiating this gives the action of the Lie algebra.

Definition 3 If ${V}$ is a representation of ${L}$, then the dual space ${V^*}$ is a representation of ${V}$ under $\displaystyle ( Xf)(v) = - f(Xv).\ \ \ \ \ (2)$

Again, the choice comes from the theory of Lie groups, where if ${G}$ acts on ${V}$, then ${G}$ acts on ${V^*}$ by $\displaystyle (gf)(v) = f(g^{-1}v).$

So, we now know how to make ${V^*}$ and ${V \otimes W}$ into ${L}$-representations for any ${V, W}$. Thus, we can make ${V^* \otimes W = \hom(V,W)}$ into a ${L}$-module. The action is given by $\displaystyle ( Xf)(v) = X (fv) - f(Xv),\ \ \ \ \ (3)$

which follows from unwinding the definitions (1) and (2).

There is a key idea here:

Proposition 4 ${f \in \hom(V,W)}$ is left invariant by ${L}$, i.e. ${Xf = 0}$ for all ${X \in L}$, if and only if ${f}$ is an ${L}$-homomorphism.

This in turn follows from (3).

In general, if we have some representation ${V}$ of ${L}$, then an invariant element ${v \in V}$ is one annihilated by ${L}$: In the setting of Lie groups, this corresponding to being fixed by the group.

Bilinear forms

Given ${L}$-representations ${V, W}$, the set of bilinear forms ${V \times W \rightarrow k}$ is isomorphic functorially to ${(V \otimes W)^*}$ and consequently has an action of ${L}$, defined by $\displaystyle (X B)(v,w) = - B( Xv, w) - B(v, Xw),$

so an invariant bilinear form is one satisfying $\displaystyle XB = 0, \quad \mathrm{or} \quad B(Xv,w) = - B(v, Xw).$

On the ${L}$-module ${L}$ with the adjoint representation, there is a useful way of constructing invariant bilinear forms. Let ${M}$ be an ${L}$-representation, and denote for any ${X \in L}$ by ${X_M}$ the linear transformation ${X_M: M \rightarrow M}$ obtained by multiplication by ${X}$.

Define the form $\displaystyle B_M( X, Y) = Tr( x_M y_M), \quad L \times L \rightarrow k.$

This is a symmetric bilinear form.

Proposition 5 ${B_M}$ is ${L}$-invariant, where ${L}$ is given the adjoint action on itself.

Proof: This is a simple computation, using the definition ${[X,Y]_M = X_M Y_M - Y_M X_M}$: $\displaystyle B_M( (ad \ X)Y , Z) = B_M ( [X,Y], Z) = Tr( X_MY_M Z_M -Y_MX_M Z_M ).$

Similarly $\displaystyle B_M( Y, (ad \ X)Z)) = B_M( Y, [X,Z]) = Tr( Y_M X_MZ_M - Y_M Z_M X_M).$

Since ${Tr( X_M Y_M Z_M) = Tr( Y_M Z_M X_M)}$ (in general, ${Tr(AB)=Tr(BA)}$ for matrices ${A,B}$), the claim follows. $\Box$